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Question Number 159680 by cortano last updated on 20/Nov/21

$${\int }_0^{\frac{\pi }{6}}\frac{\sin x\sin (x+60°)\sin (x+120°)}{\cos 3x+\sin 3x}dx=?$$

Commented by cortano last updated on 20/Nov/21

$$let\theta =3x\Rightarrow I=\frac{1}{12}{\int }_0^{\frac{\pi }{2}}\frac{\sin \theta }{\cos \theta +\sin \theta }d\theta$$$$2I=\frac{1}{12}{\int }_0^{\frac{\pi }{2}}\frac{\sin \theta +\cos \theta }{\cos \theta +\sin \theta }d\theta$$$$I=\frac{1}{24}{\int }_0^{\frac{\pi }{2}}d\theta ={\left[\frac{\theta }{24}\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{48}$$

Commented by tounghoungko last updated on 20/Nov/21

$$\Rightarrow \sin (x+60°)\sin (x+120°)$$$$=(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x)(-\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x)$$$$=\frac{3}{4}\cos {}^{2}x-\frac{1}{4}\sin {}^{2}x=\frac{3}{4}(1-\sin {}^{2}x)-\frac{1}{4}\sin {}^{2}x$$$$=\frac{3}{4}-{\sin }^{2}x$$$$\Rightarrow \sin x(\frac{3}{4}-\sin {}^{2}x)=\frac{1}{4}(3\sin x-4\sin {}^{3}x)$$$$=\frac{1}{4}\sin 3x$$$$I=\frac{1}{4}{\int }_0^{\frac{\pi }{6}}\frac{\sin 3x}{\cos 3x+\sin 3x}dx$$$$$$

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